Work Energy And Power Question 230

Question: A force acts on a 30gm particle in such a way that the position of the particle as a function of time is given by $ x=3t-4t^{2}+t^{3} $ , where x is in metres and t is in seconds. The work done during the first 4 seconds is

Options:

A) 576mJ

B) 450mJ

C) 490mJ

D) 530mJ

Show Answer

Answer:

Correct Answer: A

Solution:

[a] $ x=3t-4t^{2}+t^{3}=\frac{dx}{dt}=3-8t+3t^{2} $

Acceleration $ =\frac{d^{2}x}{dt^{2}}=-8+6t $

Acceleration after $ 4 sec = -8 + 6\times 4 =16 m{s^{-2}} $

Displacement in $ 4 sec =3\times 4-4\times 4^{2}+4^{3}=12m $

$ \therefore $ Work = Force $ \times $ displacement $ =Mass\times acc.\times disp.=3\times 1{0^{-3}}\times 16\times 12=576mJ $



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